Question: Divide the following complex numbers. $ \dfrac{-2-2i}{-2-2i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2+2i}$ $ \dfrac{-2-2i}{-2-2i} = \dfrac{-2-2i}{-2-2i} \cdot \dfrac{{-2+2i}}{{-2+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-2-2i) \cdot (-2+2i)} {(-2-2i) \cdot (-2+2i)} = \dfrac{(-2-2i) \cdot (-2+2i)} {(-2)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-2-2i) \cdot (-2+2i)} {(-2)^2 - (-2i)^2} = $ $ \dfrac{(-2-2i) \cdot (-2+2i)} {4 + 4} = $ $ \dfrac{(-2-2i) \cdot (-2+2i)} {8} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2-2i}) \cdot ({-2+2i})} {8} = $ $ \dfrac{{-2} \cdot {(-2)} + {-2} \cdot {(-2) i} + {-2} \cdot {2 i} + {-2} \cdot {2 i^2}} {8} $ Evaluate each product of two numbers. $ \dfrac{4 + 4i - 4i - 4 i^2} {8} $ Finally, simplify the fraction. $ \dfrac{4 + 4i - 4i + 4} {8} = \dfrac{8 + 0i} {8} = 1 $